Question

求过程？第七题！

Answer 1

原式=lim 2sin（x-1）cos（x-1）/2x
=lim sin（x-1）cos（x-1）/x
=0x1/1
=0

追问

三角公式？

追答

洛必达法则

Answer 2

原式=lim(x->1)[sin(x-1)/(x-1)]²·(x-1)²/(x²-1)
=lim(x->1)[sin(x-1)/(x-1)]²·lim(x->1)(x-1)²/(x²-1)
=1×lim(x->1)（x-1）/(x+1)
=1×0
=0